Answer:
The glider's position at t=0.300s is 8.1 cm
Explanation:
Given;
maximum speed of the glider, [tex]v_{max}[/tex] = 30.0 cm/s = 0.3 m/s
period of the oscillation, T = 2.4 s
Maximum speed of oscillation is given by;
[tex]v_{max}[/tex] = ωA
Where;
A is the amplitude of the oscillation
ω is the angular speed
[tex]\omega = \frac{2\pi}{T} \\\\\omega =\frac{2\pi}{2.4}\\\\ \omega = 0.833 \pi[/tex]
The amplitude of the oscillation is given by;
[tex]A = \frac{v_{max}}{\omega} \\\\ A = \frac{0.3}{0.833\pi}\\\\ A = 0.1146 \ m[/tex]
The position of the particle is given by;
x(t) = A cosωt
x(0.3) = (0.1146)cos(0.833π x 0.3)
x(0.3) = (0.1146)cos(0.833π x 0.3)
x(0.3) = (0.1146)cos(45)
x(0.3) = 0.081 m
x(0.3) = 8.1 cm
Therefore, the glider's position at t=0.300s is 8.1 cm
